EE705(B) HIGH VOLTAGE D.C. TRANSMISSION...Graetz circuits (simpli ed analysis only) - with and without overlap. Analysis of 2 & 3 valve conduction mode and 3 & 4 valve conduction mode

EE705(B)

HIGH VOLTAGE D.C. TRANSMISSION

AnithKrishnan

October 26, 2013

Note

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ANITH [email protected]

1

Syllabus

Module I

DC power transmission - comparison of AC and DC transmission - Economics of Power trans-mission - Technical performance - Advantages and disadvantages of DC transmission - Reliability- Application of DC transmission. Types of DC links. Converter Station - Converter Units.Planning for HVDC transmission - Choice of voltage level - Modern trends in DC transmission.

Thyristor valve - valve firing - valve design consideration - Grading and damper circuit design- valve protection. Valve tests - Dielectrical and operational tests.

Module II

HVDC Converters - Analysis, Pulse number. Choice of Converter configuration - valve rating -transformer rating. Graetz circuits (simplified analysis only) - with and without overlap. Analysisof 2 & 3 valve conduction mode and 3 & 4 valve conduction mode. Converter bridge characteristics- Rectifier and Inverter characteristics of a 6 pulse and 12 pulse converter.

Module III

Principles of DC link control. Converter control characteristics - modification of control charac-teristics - system control hierarchy - firing angle control - individual phase control - equidistantpulse control. Current and extinction angle control. Starting and stopping of Dc link - powercontrol. Stabilization of AC ties. Converter faults and protection - Converter faults, protectionagainst over current and voltages in a converter station - Surge arrestor - protection against overvoltage.

Module IV

Smoothing reactors - DC lines - DC line insulators - DC breakers - basic concept, characteristics,types and applications. Sources of reactive power - static VAR systems - Thyristor controlledreactor - Types of AC filters (Basic concept only) - DC filters - Carrier frequency and RI noise.Multiterminal DC system - Potential. Application and type. Modeling of DC network.

Simulation of HVDC system - system simulation - philosophy and tools only.

2

Contents

I Notes 5

1 Introduction 61.1 Types of HVDC links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2 HVDC Transmission system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Comparison of HVDC and HVAC transmission systems . . . . . . . . . . . . . . . 7

1.3.1 Economic Aspects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3.2 Technical Performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.3.3 Reliability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4 Applications of HVDC transmission system . . . . . . . . . . . . . . . . . . . . . . 11

2 HVDC Converters 122.1 Choice of converter configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.1.1 Pulse number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.1.2 Valve utilization factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.1.3Vd0

E. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.1.4 TUF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Analysis of Converters - Graetz Circuit 153.1 Assumptions made for analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.2 Modes of operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.3 Preliminary data for analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.4 Mode 1: Without overlap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.4.1 DC output voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.4.2 Power factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.4.3 Transformer current on the secondary side . . . . . . . . . . . . . . . . . . . 173.4.4 Transformer rating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.4.5 Power rating of the valve . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.5 With overlap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.5.1 2 or 3 valve conduction mode . . . . . . . . . . . . . . . . . . . . . . . . . . 183.5.2 Equivalent Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.5.3 3 or 4 valve conduction mode . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4 Control of HVDC Systems 254.1 Desired Control Features . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.2 Principles of DC link control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.3 Control Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.4 Equivalent Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.5 Firing Angle Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.5.1 Individual Phase Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.5.2 Equidistant Pulse Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.6 Current Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.7 Extinction Angle Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3

CONTENTS 4

4.8 Starting and Stopping of dc links . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.8.1 Start-up of dc link . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

5 Faults and Protection 295.1 DC Smoothing Reactors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

5.1.1 Functions of smoothing reactors . . . . . . . . . . . . . . . . . . . . . . . . 295.1.2 Disadvantages and Limitations . . . . . . . . . . . . . . . . . . . . . . . . . 295.1.3 Principle of smoothing effect . . . . . . . . . . . . . . . . . . . . . . . . . . 305.1.4 Criterion for choice of smoothing reactor . . . . . . . . . . . . . . . . . . . . 30

6 Harmonics and Filters 316.1 Harmonics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316.2 Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

6.2.1 Type of connection: Series or Shunt . . . . . . . . . . . . . . . . . . . . . . 316.3 AC filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

6.3.1 Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326.4 DC filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

7 Reactive Power Compensation 337.1 Sources of reactive power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

7.1.1 Synchronous Condensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

II University Exam Questions and Answers 35

8 Module I 368.1 Short Answer Questions (5 marks) . . . . . . . . . . . . . . . . . . . . . . . . . . . 368.2 Long Answer Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

9 Module II 389.1 Short Answer Questions (5 marks) . . . . . . . . . . . . . . . . . . . . . . . . . . . 389.2 Long Answer Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

10 Module III 4010.1 Short Answer Questions (5 marks) . . . . . . . . . . . . . . . . . . . . . . . . . . . 4010.2 Long Answer Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

11 Module IV 4311.1 Short Answer Questions (5 marks) . . . . . . . . . . . . . . . . . . . . . . . . . . . 4311.2 Long Answer Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

A Formulae 47

B Derivations 48B.1 Mode 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

College of Engineering, Kidangoor 2006 Scheme

Part I

Notes

5

Chapter 1

Introduction

1.1 Types of HVDC links

1. Monopolar linkThe monopolar link had got one conductor, usually of negative polarity, and ground or seareturn.

2. Bipolar linkThe bipolar link has two conductors one positive and the other negative. Each terminal hastwo converters of equal rated voltages in series on the dc side. The neutral points (junctionsbetween converters) are grounded at one or both ends. If both neutrals are grounded, thetwo poles can also work independently. Normally they operate at equal current; then thereis no ground current. In the event of a fault on one conductor, the other conductor withground return can carry up to half of the rated load.

3. Homopolar linkThe homopolar link has two or more conductors all having the same polarity, usually neg-ative, and always operates with ground return. In the event o f a fault on one conductor,the entire converter is available for connection to the remaining conductor/s, which, havingsome overload capability can carry more than half of the rated power, at the expense ofincreased line loss. However this is not possible in a bipolar system due to the use of gradedinsulation for negative and positive poles. In this respect, a homopolar line is preferablewhere continuous ground current not deemed objectionable. An additional minor advantageis the low power loss due to corona. Negative polarity is preferred on overhead lines becauseof its smaller radio interference.

6

CHAPTER 1. INTRODUCTION 7

1.2 HVDC Transmission system

Figure 1.1: A typical bipolar HVDC Transmission scheme

Essential components of HVDC systems are:

1. 6/12 pulse converters

2. Converter transformer with suitable ratio and tap changing

3. Filters both on the ac side and on the dc side to reduce the harmonics generated. There aretwo types of filters - tuned (on ac side) and band pass

4. A smoothing reactor to smoothen the DC current and other possible transient over-currents.

5. Shunt capacitor to complement the reactive power generated by the converters.

6. DC transmission line/cable

1.3 Comparison of HVDC and HVAC transmission systems

1.3.1 Economic Aspects

1. For a given power level, dc line requires less Right-of-Way (RoW), simpler and cheapertowers and reduced conductor and insulator costs. The power loss is also reduced with dcas there are only two conductors. The absence of skin effect is with dc is also beneficial inreducing power losses marginally. The dielectric loss in case of power cabled is also very lessin case of dc transmission.

Let us assume that an ac line and a dc line using the same conductors and insulators arebuilt. Assume that in each case, the current is limited by temperature rise. Then the dccurrent equals the rms current.

Assume also that the insulators withstand the same crest voltage to ground in each case.Then the direct voltage is

√2 times the rms alternating voltage.

The dc power per conductor ispd = VdId (1.1)

and the ac power per conductor is

pa = VaIa cosφ (1.2)



where Id and Ia are the currents per conductor, Vd and Va the conductor-to-conductorground voltages, and cosφ the power factor. The ratio is

pdpa

=VdId

VaIa cosφ(1.3)

Take cosφ = 0.945 and pd

pa= 1.5. Now compare a three phase, three conductor ac line with

a bipolar two conductor dc line. The power capabilities of the respective circuits are

Pd = 2pd

Pa = 3pa

and the ratio is

Pd

Pa=

2pd3pa

= 1

Both lines carry the same power. The dc line is simpler and cheaper, having two conductorsinstead of three.

2. The corona loss on dc conductors tend to be less significant than for ac and this also leadsto the choice of economic size of conductors with dc transmission. The other factors thatinfluence the line costs are the costs of compensation and terminal equipment. DC lines donot require compensation but the terminal equipment costs are increased due to the presenceof converters and filters.

3. Length of Transmission line: 3. The variation of cost of ac and dc transmission is shownin figure 1.2. DC transmission is economical only beyond a certain distance d∗, called thebreakeven distance. The breakeven distance also varies with the power transmitted.

Figure 1.2: Variation of cost of ac and dc transmission vs. distance

1.3.2 Technical Performance

1. Current Limit: The temperature of a conductor must be limited in order to avoid damageto the conductor itself (permanently increased sag) or, in case of a cable, to the insulationin contact with it. Hence the current in the conductor must be limited in accordance withits duration and the ambient temperature. The ac resistance of a conductor is higher thanits dc resistance because of skin effect. For a given conductor, the current carrying capacityof HVDC system is more than the HVAC system for the same voltage level.

2. Reactive power and voltage regulation (or voltage control): On long EHVAC over-head lines and on much shorter ac cables, the production and consumption of reactive powerby the line itself constitutes a serious problem. On a line having series inductance L and



shunt capacitance C per unit length and operating voltage V and current I, the line producesreactive power

QC = ωCV 2 (1.4)

and consumes reactive power

QL = ωLI2 (1.5)

per unit length. The reactive power produced by the line equals that consumed by it, withno net production or consumption, if

ωCV 2 = ωLI2 (1.6)

hence ifV

I=

[L

C

]1/2= Zs (1.7)

In this case, the load impedance has the value Zs, known as the surge impedance of the line.The power carried by the line so loaded is

Pn = V I =V 2

Zs(1.8)

and is called the surge impedance loading (SIL) or natural load. It is independent of distanceand depends mainly on the voltage.

Figure 1.3: Variation of voltage along the transmission line (uncompensated)

For stipulated constant voltages at the sending end and receiving end, the voltage profilevaries with line loading. The voltage profile is relatively flat only for a fixed level of powertransfer corresponding to SIL (P = Pn). Under light loads (P < Pn), the mid-point voltagemay exceed the upper permissible voltage limit and at high loads (P > Pn), the voltage maydip below the lower permissible limit.

Another limitation of long lines is the high voltage at an open end (the Ferranti effect). Thisis important when a line is being put into service by first connecting one end of it to themain ac system, for it is not feasible to close both ends at exactly the same moment.

The maintenance of constant voltages at the two ends requires reactive power control frominductive to capacitive (mid-point shunt, inductive or capacitive or static Var compensators,compensation) as the line loading is increased. The reactive power requirements also in-crease with increase in line lengths. Although DC converter stations with line commutatedconverters require reactive power (drawn from the ac side, usually shunt capacitors are pro-vided for this) related to the line loadings (independent of line length), the line itself doesnot require reactive power, and the voltage drop on the line itself is merely the resistivedrop. For distances more than 400km the total reactive power requirement of HVDC systemis less than that of HVAC system, for the same power to be transmitted.



In the case of cables (submarine or underground), they are always operated at a load muchbelow the SIL in order to avoid over-heating. Consequently, the reactive power produced bycharging the shunt capacitance greatly exceeds that consumed by the series inductance. In a50Hz power cable of length approximately 40km, the charging current alone equals the ratedcurrent. Shunt compensation theoretically would solve this problem, but practically it isvery difficult to lay and maintain (or repair) such cables. DC cables have no such limitation.

3. Stability: The stability of an ac system is its ability to operate with all synchronousmachines in synchronism. If a long ac line is loaded to a certain value, known as its steady-state stability limit, the synchronous machines at the sending end will go out of synchronismwith those at the receiving end. This will in turn give rise to objectionable fluctuations involtage.

Even if a line is operated below its steady-state limit, the machines at the sending andreceiving ends may lose synchronism after some large disturbance, notably a short circuit,unless the line is operated below its transient stability limit, which is always lower thanthe steady-state limit. Practically speaking, the steady-state stability limit is the transientstability limit for a very small disturbance.

The problem of stability or synchronous operation constitutes the most serious limitation ofa long ac transmission system. The power transmitted from the sending end to the receivingend in a lossless ac power system is given by

P =E1E2

Xsin δ (1.9)

where E1 and E2 are the voltages of sending end and receiving end of the power system,δ is the phase difference between these voltages (power angle or internal angle), and X isthe reactance of the connecting network. Thus the maximum power that can be transferredover a long transmission line is inversely proportional to the line reactance. Whereas in DCtransmission, the power transmitted is limited only by the current carrying capacity of theconductors. A dc transmission link in itself has no stability problem.

Figure 1.4: Power transfer capability vs. distance

4. Circuit Breakers: AC circuit breakers take advantage of the current zeros that occur twiceper cycle. They are designed to increase the breakdown strength of the arc path betweencontacts so rapidly that the arc does not re-strike. DC circuit breakers do not have thisnatural advantage and therefore have to force the current to zero. In simple two-terminal dctransmission, the lack of dc circuit breakers have not been felt, because faults on the dc lineor in the converter are cleared by using the control grids of the converter valves to block thecurrent temporarily.

5. Ground Return: A two conductor bipolar dc line is more reliable than a three conductorac line, because, in the event of a fault on one conductor, the other conductor can continue



to operate with ground return during the period required for operating the fault. Theoperation of an ac line with ground return is not feasible on account of high impedanceof such a circuit and the telephonic interference caused by such an operation. The groundreturn in dc transmission is objectionable only when buried metallic structures (such aspipes) are present and are subject to corrosion with DC current flow.

6. Problems of ac interconnection: When two power systems are connected through acties (synchronous interconnection), the automatic generation control of both systems has tobe coordinated using tie-line power and frequency signals. Even with coordinated control,of interconnected systems, the operation of ac ties can be problematic due to:

(a) presence of large power oscillations which can lead to frequent tripping

(b) increase in fault levels

(c) transmission of disturbances from one system to the other.

The controllability of power flow in dc lines eliminates all the above problems. In addition,for asynchronous dc ties, there is no need of coordinated control. It is obvious that twosystems which have different nominal frequencies cannot be interconnected directly with acties and require the use of dc ties.

1.3.3 Reliability

There are two measures of overall system reliability.

1. Energy Availability: Is defined as

Energy Availability = 100×(1− equivalent outage

total time

)%

where equivalent outage time is the product of the actual outage time and the fraction ofsystem capacity lost due to outage.

2. Transient Reliability: This is a factor specifying the performance of HVDC systems duringrecordable faults on the associated AC systems.

Transient Reliability =No. of times HVDC system performed as designed

no. recordable ac faults

Recordable ac system faults are those faults which cause one or more AC bus phase voltageto drop below 90

A bipolar dc line is as reliable as a double circuit ac line with the same power transfer capability.This is because of the fact that failure of one pole does not affect the operation of the other pole(with ground return).

1.4 Applications of HVDC transmission system

1. Long distance bulk power transmission

2. Power transmission using underground or underwater cables (for distance > 32km)

3. Asynchronous interconnection of AC systems operating at different frequencies or whereindependent control of systems is desired

4. Control and stabilisation of power flows in AC ties in an integrated power system


Chapter 2

HVDC Converters

2.1 Choice of converter configuration

The choice of converter configuration is made on the following requirements:

1. High pulse number

2. Valve utilization factor, PIVVd0

should be as low as possible

3. Vd0

E should be as high as possible

4. Transformer Utilization Factor (TUF) should be as low as possible

2.1.1 Pulse number

The number of pulsations or ripples of dc voltage per cycle of ac voltage is known as pulse numberof a converter.

If q is the number of valves in a commutation group1 and r of these are connected in paralleland s of them are connected in series, the pulse number is given by

p = qrs (2.1)

High pulse numbers will result in lesser harmonics generated as the characteristic harmonicsgenerated on the ac side is given by h = np±1 and on the dc side is given by h = np, where n is aninteger. Thus low order harmonics, which have comparatively higher magnitudes are eliminatedwhen p is high.

Higher the pulse number, higher will be the average output dc voltage level per cycle of acwave.

Since it is easy to design filters (low pass filters) that eliminate higher order harmonics, highpulse number is preferred.

Higher pulse number also results in better utilisation of converter transformer and valves.The pulse number cannot be increased indiscriminately as the number of valves in the circuit

increases for increase in pulse number. This will increase the overall cost of the converter station.

2.1.2 Valve utilization factor

The valve utilization factor (VUF) is given by

V UF =PIV

Vd0(2.2)

1A group of valves in which only one valve conducts at any instant, neglecting overlap.

12

CHAPTER 2. HVDC CONVERTERS 13

where PIV is the peak inverse voltage, the maximum reverse bias voltage that the valve is subjectedto and Vd0 is average output dc voltage when there is no delay in firing the valves i.e. α = 0. Itis always good when PIV is less. This is one reason why we go for bridge converter circuits. Theconverter configuration should be so chosen such that Vd0 is maximum.

If there are q valves in a commutation group, each valve will conduct for a period of 2π/q. Theaverage maximum dc output voltage of the converter is

Vd0 =1

2π/q

∫ π/q

−π/q

Em cosωt dωt (2.3)

=q

πEm sin

π

q(2.4)

where Em =√2E, is the peak value of the input alternating voltage and ω = 2πf is the supply

frequency in rad/sec. If there are s number of series connections, then the dc voltage is

Vd0 =sq

πEm sin

π

q(2.5)

The expression for PIV is

PIV =

2Em if q is even

2Em cosπ

2qif q is odd (2.6)

and therefore,

PIV

Vd0=

2π

sq sinπ

q

if q is even

π

sq sinπ

2q

if q is odd(2.7)

2.1.3Vd0

EThis ratio should be high as we prefer a very minimal voltage drop in the conversion circuit. UsingEq. (2.5), we get

Vd0

E=

√2qs

πsin

π

q(2.8)

2.1.4 TUF

TUF is defined as the ratio of transformer rating (of the valve side) to the dc power output. Thisvalue should be as low as possible. The current rating of the transformer is given by

It =Idr√q

(2.9)

where Id is the dc current through the link. The transformer rating will be

Pt = pEm√2It = p

EmIdr√2q

(2.10)

Thus TUF can be calculated as

TUF =Pt

Vd0Id=

π√2q sin π

q

(2.11)

An important point to be noted here is that the value of TUF is dependant on the value q only.


CHAPTER 2. HVDC CONVERTERS 14

Sl. No. q r s PIV/Vd0 Vd0/E TUF1 2 1 3 1.047 2.700 1.5712 2 3 1 3.142 0.900 1.5713 3 1 2 1.047 2.340 1.4814 3 2 1 2.094 1.169 1.4815 6 1 1 2.094 1.350 1.814

Table 2.1: Converter Configuration - 6 pulse

From the table it can be seen that sl. no. 1 and 3 are suitable, but TUF is better for sl. no.3. Since 3 phase supply is used, the ease of connection is an added advantage in using sl. no. 3.


Chapter 3

Analysis of Converters - GraetzCircuit

3.1 Assumptions made for analysis

The following assumptions are made to make analysis simpler.

1. The ac power source is a balanced sinusoidal source with constant amplitude and frequencyand has no impedance.

2. The dc current is ripple free i.e. the dc load has infinte inductance.

3. Valves have zero forward resistance when conducting and infinite resistance when not con-ducting.

4. Valves are ignited at equal intervals of 60.

3.2 Modes of operation

Depending on the number of valves conducting at any instant, there are different modes of oper-ation. Ideally, only two valves conducts (one from upper commutation group and one from thelower commutation group) at any given time and it is assumed that the current transfer fromoutgoing valve to the incoming valve is instantaneous. This does not happen in real time circuitsi.e. there will be more than two valves conducting simultaneously and this phenomenon is calledoverlap. The phenomenon of overlap is caused mainly due to

1. leakage reactance of the transformer, and

2. the unsymmetric firing of valves

When there is overlap, the current transfer is not instantaneous.Depending on the period of overlap, denoted by u, there are different modes of operation. They

are

Mode 1 u = 0, 2 valve conduction modeWhen u = 0, there is no overlap. This is the ideal mode of working of a converter circuit.

Mode 2 u < 60, 2 or 3 valve conduction modeThis is the most practical mode of working of a converter circuit.

Mode 3 u = 60, 3 valve conduction mode

Mode 4 u > 60, 3 or 4 valve conduction mode

15

CHAPTER 3. ANALYSIS OF CONVERTERS - GRAETZ CIRCUIT 16

3.3 Preliminary data for analysis

Before we move on to the analysis, we need to define the voltage phasors and the initial conditionof the circuit.

Taking eba as the reference voltage, the other voltages can be derived as

eba =√3Em sinωt (3.1)

ea = Em sin

(ωt+

5π

6

)(3.2)

eb = Em sin(ωt+

π

6

)(3.3)

ec = Em sin(ωt− π

2

)(3.4)

ecb =√3Em sin

(ωt− 2π

3

)(3.5)

eac =√3Em sin

(ωt+

2π

3

)(3.6)

ebc =√3Em sin

(ωt+

π

3

)(3.7)

A valve can be triggered when its anode to cathode voltage is positive. It can be delayed by angleα.

3.4 Mode 1: Without overlap

In this mode of operation u = 0 and only two valves conduct at any instant and the currenttransfer is instantaneous.

3.4.1 DC output voltage

Since each pair of valves conducts for a period of 60, the voltage ripples will be symmetric overeach period. Thus the average dc voltage can be calculated as

Vd =1

π/3

∫ π/3+α

α

ebc dωt (3.8)

=3√3Em

π

∫ π/3+α

α

sin(ωt+

π

3

)dωt (3.9)

=3√3Em

πcosα (3.10)

= Vd0 cosα (3.11)

where Vd0 is called the ideal no-load direct voltage and is defined as

Vd0 =3√3Em

π(3.12)

The polarity of Vd depends upon the value of α. Thus Vd varies from Vd0 to −Vd0, which impliesthat the same converter can act as a rectifier or inverter depending upon the value of α. The dcoutput voltage is maximum when α = 0 and zero when α = 90.

3.4.2 Power factor

With losses in the converter neglected, the ac power must be equal to the dc power, i.e.

3ELNIL1 cosφ = VdId = Vd0Id cosα (3.13)



where IL1 is the rms value of the fundamental component of the alternating line current. The linecurrent has the waveform as shown in figure and consists of positive and negative rectangularpulses of magnitude of Id and width 2π/3. This shape is independent on α as long as there is nooverlap. The rms value of the fundamental component can be found out using Fourier analysis

IL1 =1√2

[2

π

∫ +π/3

−π/3

Id cos θ dθ

](3.14)

=

√2

πId [sin θ]

π/3−π/3 (3.15)

=

√6

πId (3.16)

Substituting Eq.(3.16) and Eq.(3.12) into Eq.(3.13), we get

cosφ = cosα (3.17)

The above equations shows that power factor decreases when firing angle is increased and thusthe converter requires more reactive power.

3.4.3 Transformer current on the secondary side

The rms value of the overall current is given by

I =

[1

2π

∫ 2π

0

I2d dωt

] 12

(3.18)

=Id2π

[∫ 5π/6

π/6

1 dωt+

∫ 11π/6

7π/6

1 dωt

] 12

(3.19)

=

√2

3Id (3.20)

The rms value of the hth harmonic current Ih is given by

Ih =I1rms

h(3.21)

where 1rms is the rms value of the fundamental component, h is the order of harmonics given by

h = np± 1

where p is the number of pulses, n is an integer.

3.4.4 Transformer rating

Transformer rating (denoted as T ) in Volt-Amperes can be calculated using

T =π

3Vd0Id (3.22)

3.4.5 Power rating of the valve

3.5 With overlap

Due to the inductance of the transformer, the current in it can vary only at a finite rate andthus the transfer of current from one phase to another phase will take finite time, called thecommutation time or overlap time.



3.5.1 2 or 3 valve conduction mode

DC Voltage

When valves 1 and 3 are conducting (i.e. during the period of overlap), lines a and b are shortcircuited.

Figure 3.1: 3 valve conduction

The voltage Va can be calculated as follows

Va = ea − LdIadt

(3.23)

Va = eb − LdIbdt

(3.24)

Adding the above two equations

2Va = ea + eb − Ld (Ia + Ib)

dt(3.25)

But Ia + Ib = Id, the dc current and is assumed to be a constant. Therefore

2Va = ea + eb − LdIddt

(3.26)

= ea + eb

(∵ dId

dt= 0

)∴ Va =

ea + eb2

(3.27)

For a balanced 3 phase supply ea + eb + ec = 0, and hence

Va =ea + eb

2= −ec

2(3.28)

The average dc terminal voltage can be obtained as

Vd = Va − ec =

ea+eb

2 − ec2 = −3ec

2 α ≤ ωt ≤ u+ αeb − ec = ebc u+ α ≤ ωt ≤ π

3 + α(3.29)

Therefore,

Vd =3

π

[∫ u+α

α

−1.5ec dωt+

∫ π3 +α

u+α

ebc dωt

]

=3√3Em

2π[cosα+ cos (α+ u)]

=Vd0

2[cosα+ cos (α+ u)] (3.30)



DC Current

From Eq. (3.23) and (3.24),

ea − LdIadt

= eb − LdIbdt

(3.31)

L

[dIbdt

− dIadt

]= eb − ea (3.32)

Since Ia = Id − Ib,

dIadt

= −dIbdt

(3.33)

Therefore Eq.(3.32) becomes

2LdIbdt

= eba =√3Em sin (ωt) (3.34)

Integrating on both sides

Ib =−√3Em

2ωLcos (ωt) +A (3.35)

where A is the integration constant. A can be calculated from the initial conditions. At ωt = α,the current through valve 3 is zero i.e. I3 = 0.

∴ A =

√3Em

2ωLcosα (3.36)

Eq.(3.35) now becomes

Ib =−√3Em

2ωLcos (ωt) +

√3Em

2ωLcosα (3.37)

=

√3Em

2ωL[cosα− cosωt] (3.38)

Eq.(3.38) is valid only during the commutation period i.e. from α to α + u. The current of theincoming valve during commutation, consists of a constant (dc) term and a sinusoidal term. Thesinusoidal term lags behind the commutating voltage by 90, as it should in a purely inductivecircuit. The equation for Ia can be deduced from Ia = Id − Ib. At the end of commutation i.e. atωt = α+ u, Ib = Id. Therefore

Id =

√3Em

2ωL[cosα− cos (α+ u)] (3.39)

Let Is =

√3Em2ωL , therefore

Id = Is [cosα− cos (α+ u)] (3.40)

Power Factor

3ELNIL1 cosφ = VdId (3.41)

From eq.(3.30),

Vd =3√6ELN

2π[cosα+ cos (α+ u)] (3.42)



Therefore,

3ELNIL1 cosφ =3√6ELN

2π[cosα+ cos (α+ u)] Id (3.43)

IL1 cosφ =

[√6

πId

] [cosα+ cos (α+ u)

2

](3.44)

Using the following approximation,

IL1 u√6

πId (3.45)

cosφ =cosα+ cos (α+ u)

2(3.46)

3.5.2 Equivalent Circuit

From eq.(3.39),

cos (α+ u) = cosα− 2ωLId√3Em

Substituting the above equation into eq.(3.30) and eliminating cos (α+ u),

Vd =3√3Em

2π

[2 cosα− 2ωLId√

3Em

](3.47)

=3√3Em

πcosα− 3ωL

πId (3.48)

= Vd0 cosα− 3ωL

πId (3.49)

Let Rc = 3ωLπ , be defined as the equivalent commutation resistance. Although it is called

as resistance, it does not consume power and is used only for representing a voltage drop due tooverlap. Therefore

Vdr = Vd0r cosα−RcId (3.50)

Eq.(3.50) is valid for the rectifier and the same equation is modified so that it represents theinverter. Let π−α = β and π− (α+ u) = γ, where γ is called the extinction angle and β is calledthe angle of advance.

Eq.(3.30) can now be modified as

Vdi = −Vd0i

2(cos (π − β) + cos (π − γ))

The voltage is indicated as negative, since the representation is for the inverter side. Therefore,

Vdi =Vd0i

2(cosβ + cos γ) (3.51)

From eq.(3.12),

Vdi = −Vd0i cos (π − β) +RcId

Vdi = Vd0i cosβ +RcId (3.52)

Modifying eq.(3.51), we get

cosβ = − cos γ +2Vdi

Vd0i(3.53)



Substituting eq.(3.53) into eq.(3.52)

Vdi = Vd0i

[− cos γ +

2Vdi

Vd0i

]+RcId

= −Vd0i cos γ + 2Vdi +RcId

∴ Vdi = Vd0i cos γ −RcId (3.54)

Figure 3.2: Equivalent circuit of an HVDC link.

3.5.3 3 or 4 valve conduction mode

Operation of bridge converter with overlap angle u > 60 is abnormal, being encountered onlyunder overload, dc short circuit, or low alternating voltage. However, it is self curing, i.e. it willcontinue conducting in 3 valve mode after overlap period. When 4 valves are conducting, theyconstitute a 3 phase short circuit on the ac side and a pole-to-pole short circuit on the dc side.When 3 valves are conducting, they constitute a line-to-line short circuit on the ac side (as seenin the previous section).

In this mode of operation, when 4 valves are conducting, the direct voltage is zero, and when3 valves are conducting, the output is an arc of a sine wave of amplitude 1.5Em.

Figure 3.3: 4 valve conduction

DC Voltage

The average dc terminal voltage can be obtained as

Vd =

0 α ≤ ωt ≤ α+ u− π

3−1.5ec α+ u− π

3 ≤ ωt ≤ α+ π3

(3.55)



Therefore the output dc voltage is

Vd =3

π

∫ α+π3

α+u−π3

−1.5ec (3.56)

= − 9

2π

∫ α+π3

α+u−π3

Em sin(ωt− π

2

)dωt

=9

2πEm

∫ α+π3

α+u−π3

cosωt dωt

=9

2πEm

[sinωt

]α+π3

α+u−π3

=9

2πEm

[cos

(ωt− π

2

)]α+π3

α+u−π3

=9

2πEm

[cos

(α− π

6

)− cos

(α+ u− 5π

6

)]=

9

2πEm

[cos

(α− π

6

)+ cos

(α+ u+

π

6

)]=

√3

2Vd0

[cos

(α− π

6

)+ cos

(α+ u+

π

6

)](3.57)

DC Current

When valves 6,1,2,3 are conducting,

ea − LdIadt

= eb − LdIbdt

(3.58)

ea − LdIadt

= ec − LdIcdt

(3.59)

Using the above two equations, we get

ea − eb = LdIadt

− LdIbdt

(3.60)

ea − ec = LdIadt

− LdIcdt

(3.61)

Adding eq.(3.60) and eq.(3.61)

2ea − eb − ec = 2LdIadt

− LdIbdt

− LdIcdt

3ea = 2LdIadt

− LdIbdt

− LdIcdt

(∵ ea + eb + ec = 0)

Since, Ia + Ib = Id = −Ic, we get

3ea = 3LdIadt

∴ dIadt

=eaL

(3.62)

Considering the entire system,

dIadt

=eaL

=dI1dt

= −dI3dt

=Em

Lsin

(ωt+

5π

6

)(3.63)



Integrating on both sides,

I3 =

∫ ωt

α

Em

Lsin

(ωt+

5π

6

)dωt

=Em

ωL

[cos

(ωt+

5π

6

)− cos

(α+

5π

6

)]=

Em

ωL

[− cos

(ωt− π

6

)+ cos

(α− π

6

)](3.64)

The valves 6,1,2 & 3 will conduct during the period α ≤ ωt ≤ α+u− π3 . Therefore at ωt = α+u− π

3 ,the current through valve 3 will be

I3 =Em

ωL

[− cos

(α+ u− π

2

)+ cos

(α− π

6

)](3.65)

At ωt = α+ u− π3 , valve 6 will turn off completely and the valves 1, 2, & 3 will conduct (3 valve

conduction). Current through valve 3 during this period can be found by integrating the current(see eq.(3.34)) during this period and adding it to the initial current (eq.(3.65)).

I3 =

∫ ωt

u+α−π3

√3Em

2Lsinωt dωt+

Em

ωL

[− cos

(α+ u− π

2

)+ cos

(α− π

6

)]=

√3Em

2ωL

[− cosωt+ cos

(u+ α− π

3

)]+

Em

ωL

[− cos

(α+ u− π

2

)+ cos

(α− π

6

)]Valves 1, 2 & 3 will conduct during the period α+ u+ π

3 ≤ ωt ≤ α+ π3 . Therefore, at ωt = α+ π

3

I3 =

√3Em

2ωL

[cos

(u+ α− π

3

)− cos

(α+

π

3

)]+

Em

ωL

[− cos

(α+ u− π

2

)+ cos

(α− π

6

)]=

Em

ωL

[cos

(α− π

6

)− cos

(α+ u− π

2

)+

√3

2cos

(α+ u− π

3

)−

√3

2cos

(α+

π

3

)]

By vector addition,

− cos(α+ u− π

2

)+

√3

2cos

(α+ u− π

3

)=

1

2cos

(α+ u+

π

6

)(3.66)

∴ I3 =Em

ωL

[cos

(α− π

6

)+

1

2cos

(α+ u+

π

6

)−

√3

2cos

(α+

π

3

)](3.67)

At ωt = α + π3 , valve 4 is fired. Now, 4 valves (1, 2, 3 & 4) will conduct during the period

α+ π3 ≤ ωt ≤ α+ u. Analysis is done in a similar way as was done when valves 6, 1, 2 & 3 were

conducting.

ea − LdIadt

= eb − LdIbdt

= ec − LdIcdt

Combining the above equations.

2eb − ea − ec = 2LdIbdt

− LdIadt

− LdIcdt

∴ eb = LdIbdt

= LdI3dt

= Em sin(ωt+

π

6

)(3.68)



The equation for I3 can now be written as

I3 =

∫ ωt

α+π3

Em

Lsin

(ωt+

π

6

)dωt

+Em

ωL

[cos

(α− π

6

)+

1

2cos

(α+ u+

π

6

)−

√3

2cos

(α+

π

3

)]

=Em

ωL

[cos

(α+

π

2

)− cos

(ωt+

π

6

)]+

Em

ωL

[cos

(α− π

6

)+

1

2cos

(α+ u+

π

6

)−

√3

2cos

(α+

π

3

)]

At ωt = α+ u

Id = I3

=Em

ωL

[cos

(α+

π

2

)− cos

(α+ u+

π

6

)+ cos

(α− π

6

)+

1

2cos

(α+ u+

π

6

)−

√3

2cos

(α+

π

3

)]

=Em

ωL

[1

2cos

(α− π

6

)− 1

2cos

(α+ u+

π

6

)]Finally,

Id =Em

2ωL

[cos

(α− π

6

)− cos

(α+ u+

π

6

)](3.69)


Chapter 4

Control of HVDC Systems

4.1 Desired Control Features

1. The control system should not be sensitive to normal variation in voltage and frequency ofthe ac supply system.

2. Control should be fast, reliable and easy (simple) to implement.

3. It should have continuous operating range from full rectification to full inversion.

4. Control should be such that, it should require less reactive power (in order to have a goodpower factor).

5. Under steady state conditions, the values must be fired symmetrically.

6. The control should be able to limit the maximum current so as to avoid damage to valvesand other current carrying devices. It should also limit the fluctuations of current due tofluctuation of alternating voltage.

7. Power can be controlled independently and smoothly which can be done by controlling thecurrent and/or the voltage simultaneously in the link.

8. Control should be such that it can be used for protection of line and converter

• prevention of commutation failure of inverter

• prevention of arc back of the rectifier valves

4.2 Principles of DC link control

Consider the steady state equivalent circuit of a 2 terminal dc link shown in figure. This is basedon the assumption that all the series connected bridges in both poles of a converter station areidentical and have same delay angles. Also, the number of series connected bridges (nb) in bothstations (rectifier and inverter) are the same. Therefore the voltage source at the rectifier side canbe expressed as

Vdrect = nbVd0r cosα (4.1)

and that of the inverter side as

Vdinv = nbVd0i cos γ (4.2)

25

CHAPTER 4. CONTROL OF HVDC SYSTEMS 26

4.3 Control Characteristics

4.4 Equivalent Circuit

Id =Vd0r cosα− Vd0i cosβ

Rcr +Rl ±Rci(4.3)

for safe commutation margin, γ is used instead of β

∴ Id =Vd0r cosα− Vd0i cos γ

Rcr +Rl −Rci(4.4)

−Rci makes it difficult for the stable operation of the controller.

4.5 Firing Angle Control

There are basically two types of firing angle control schemes. They are

1. Individual Phase Control (IPC), and

2. Equidistant Pulse Control (EPC)

4.5.1 Individual Phase Control

The firing pulse generated for each valve is independent of each other and is determined by thezero crossing of the commutation voltage of the corresponding valve. There are two ways in whichthis can be accomplished. They are constant α control and inverse cosine control.

Constant α control

Since there are 6 valves, there will be six commutation voltages and six gate pulses will be generatedcorresponding to the zero crossing of these voltages. A block diagram for the control is shown inFigure 4.1.

Figure 4.1: Constant α control

The output of the ZCD is a rectangular pulse of period π. This same pulse can be used todrive the valve (after passing through a pulse amplifier). In order to control the instant at whichthis pulse reaches the valve, a delay circuit is used. The duration of delay is calculated from themagnitude of the control voltage Vc from the current controller. An example is shown in figure.

Inverse Cosine Control

4.5.2 Equidistant Pulse Control

Advantages

1. The scheme provides equal pulse spacing in the steady state.



2. Low non-characteristic harmonics when used with weak ac systems.

Drawbacks

1. During unbalanced ac voltages, it results in a lower dc voltage and power than IPC.

4.6 Current Control

4.7 Extinction Angle Control

Knowing commutation voltage and γmin, one can find β as shown below

2Ldi3dt

= eb − ea =√3Em sinωt (4.5)

2L

∫ Id

0

di3 =√3Em

∫ t2

t1

sinωt dt (4.6)

2LId = −√3Em

ω

[cosωt

]π−αω

α/ω(4.7)

2ωLId = −√3Em [cos (π − γmin)− cosα] (4.8)

2ωLId =√3Em [cos γmin + cosα] (4.9)

− cosα = cos γmin − 2ωLId√3Em

= cosβ (4.10)

From the above expression it is clear that if Id is measured and if ω, L, Em are known, the wecan calculate the β required.

4.8 Starting and Stopping of dc links

4.8.1 Start-up of dc link

HVDC links can be started by using either

1. long gate pulses (120) or

2. short gate pulses (60)

Long pulse firing

1. De-block the inverter at about γ = 90

2. De-block the rectifier at about α = 85 to establish low direct current

3. Gradually ramp up the voltage by inverter control and the current by rectifier control.

Short pulse firing

In this case, the problem of current extinction during start up is present as the valve with forwardbias is not put into conduction when the current in that transiently falls below holding current.

The starting sequence is as follows:

1. Open the bypass switch at one terminal

2. De-block that terminal and load to minimum current in the rectifier mode.

3. Open bypass switch at the second terminal and commutate current to the bypass pair.



4. Start the second terminal also in the rectifier mode

5. The inverter terminal is put into the inversion mode

6. Gradually ramp up the voltage and current

The voltage is normally raised before raising the current. This permits the insulation of theline to be checked before raising the power. The ramping of power avoids stresses on the generatorshaft. The switching surges on the line are also reduced.


Chapter 5

Faults and Protection

5.1 DC Smoothing Reactors

5.1.1 Functions of smoothing reactors

The smoothing reactor performs the following functions and is essential for the converter operationin HVDC system.

1. To prevent consequent commutation failures in inverters by reducing the rate of rise of directcurrent during commutation in one bridge when the direct voltage of another series connectedbridge collapses.

2. They reduce the incidence of commutation failure in inverters cause by dips in the ac voltageat the converter bus.

3. It smoothens the dc current ripples.

4. To reduce the harmonic voltage and current in the dc line.

5. To reduce the requirement of dc filters and ac filters.

6. To reduce the current and voltage transients, during sudden changes in dc power flow.

7. To reduce the crest of the short circuit current in the dc line, thereby reducing the stresseson the converter valve.

8. To limit the current in the valves during the converter bypass pair operation, due to thedischarge of shunt capacitances of the dc line.

5.1.2 Disadvantages and Limitations

1. High inductance of smoothing reactor on dc side results in slowing down of response ofcurrent control of HVDC transmission systems.

2. Reactor has additional losses.

3. The resonant frequency is reduced and the current stabilization control becomes difficult.

4. High stored energy causes high short circuit currents on dc side between pole bus and earth.

29

CHAPTER 5. FAULTS AND PROTECTION 30

5.1.3 Principle of smoothing effect

Smoothing reactor functions as a pure inductance. Current Id through the reactor cannot changeinstantaneously. Under changing current condition, the emf is induced in the reactor coil given by

e = Ldi

dt(5.1)

This emf opposes the rate of change of current. The current change takes finite time of the orderof a few micro or milli seconds. The above principle gives a smoothing effect to the dc waveform.

5.1.4 Criterion for choice of smoothing reactor

1. Inductance should be sufficient to ensure flow of dc current even when the delay angle α isclose to 90.

2. The inductance should smoothen the dc voltage and current waveforms to acceptable limits.The percentage of ripples should be reduced to acceptable limits.

3. During the inverter faults and dc line faults, the rate of rise of current should not exceed thelimit.

4. The inductance value should remain practically constant with variations in the direct current.

5. Inductance should be low, so that the dc system current noise (due to magnetostriction) islow.

There is little information available on the choice of the optimum size of the dc smoothing reactor.One criterion used is the Si factor, defined below.

Si =Vdn

LIdn(5.2)


Chapter 6

Harmonics and Filters

6.1 Harmonics

Harmonics other than np±1 are called as non-characteristic harmonics. They are generally of lowmagnitudes. The causes are

1. Normally valves are not fired at equal intervals due to unbalance of 3-phase supply.

2. Even balanced circuits with jitter in electronic circuitery produces non-characteristic har-monics.

3. controller action (especially CEA)

4. Interaction of characteristic harmonics and fundamental current in non-linear elements inpower system

5. saturation of transformers

6.2 Filters

Harmonics are induced into both ac and dc circuits. The purpose of using harmonic filters in accircuits are

1. to reduce harmonic voltages and currents in the ac power network to acceptable levels and

2. to provide all or part of the reactive power consumed by the converter.

Whereas in dc, filters are used to reduce harmonics in the dc line.

6.2.1 Type of connection: Series or Shunt

The filters can be connected in series or shunt or both.The series filter must carry the full current of the main circuit and must be insulated throughout

for full voltage to ground.The shunt filters can be grounded at one end and carries only the harmonic current for which

it is tuned, plus a fundamental current much smaller than that of the main circuit. Hence a shuntfilter is much cheaper than a series filter of equal effectiveness.

AC shunt filters have another advantage over series filters in that at fundamental frequency,the former supplies needed reactive power but the latter consumes it.

For the above said reasons, shunt filters are used exclsively on the ac side. AC filters are starconnected with grounded neutral.

On the dc side, the dc reactor constitutes all or part of the dc filters. The remainder as dcshunt filters.

31

CHAPTER 6. HARMONICS AND FILTERS 32

6.3 AC filters

AC filters are tuned for either a particular frequency (tuned filter) or a range of frequencies(damped filter). Tuned filters (high Q1 factor) are sharply tuned to one or two of the lowerharmonic frequencies such as fifth or seventh. Damped filters (low Q factor) offers low impedanceto a wide band of high frequencies, like for example from seventeenth and higher order. Dampedfilters are also called as high-pass filters.

6.3.1 Types

The following are the various types of ac filters that can be used:

1. Single tuned filter

2. Double tuned filter

3. High-pass filter

(a) Second order filter

(b) C type filter

The single tuned filters are designed to filter ot characteristic harmonics of single frequency.The double tuned filters are used to filter out two discrete frequencies, instead of using two singletuned filters. Their main advantages are:

1. its power loss at fundamental frequency is less

2. one inductor, instead of two is subjected to full impulse voltage.

The second order high-pass filters are designed to filter out the higher harmonics. The tuningof these filters is not critical. The losses at the fundamental frequency can be reduced by usinga C type filter where the capacitor C2 in series with L, provides a low impedance path to thefundamental component of current.

All the filter branches appear capacitive at fundamental frequency and supply reactive power.

6.4 DC filters

As stated before, the dc reactors, although designed primarily for other functions, constitute allor part of the filters on the dc side. If the dc line is in a cable, generally no additional filteringis required on the dc side, because the cable sheath and the ground or sea water adequatelyshield telephone lines from noise induced by the dc harmonics. With overhead dc lines, additionalfiltering is usually required.

The harmonics in the dc voltage contain both characteristic and non-characteristic orders(h = np). These harmonics result in current harmonics in dc lines and cause noise in telephonecircuits. The dc filter design procedure is similar to the ac filter design procedure except thedifference that reactive power of dc filter is not significant.

The effectiveness of the dc filter is judged by one of the following criteria:

1. Maximum voltage Telephone Influence Factor (TIF2) on dc high voltage bus.

2. Maximum induced noise voltage (INV) in milli volts/km in a parallel test line one kilometeraway from the HVDC line.

3. Maximum permissible noise to ground in dB in telephone lines close to HVDC lines.

1The sharpness of tuning is often expressed in terms of Q factor.2TIF is a measure of telephone interference caused by the harmoincs in the power circuit.


Chapter 7

Reactive Power Compensation

7.1 Sources of reactive power

The HVDC converts ac power to dc power. On ac side, the voltage and the current are not inphase and hence the current has got a quadrature component. Thus the converter requires reactivepower compensation for the satisfactory operation. AC system supplies active power as well asreactive power. However this is not enough. Hence additional compensation is provided on ac sideof converter by the following means:

1. ac filter capacitors

2. shunt capacitors

3. synchronous condensors

4. static VAr Sources (SVS)

These are shown schematically in figure.The reactive power consumption vary mainly with the following

1. Active power Pd and

2. Delay angle α of rectifer (and extinction angle γ of inverter)

Reactive power demand varies from 20% to 60% of the active power flow. This requiresadjustable reactive power source which can provide variable reactive power as demanded. Forslow variations in the load, switched capacitors or filters can provide some control. However,this is discrete type of control and can result in voltage flicker unless the size of the unit, whichis switched, is made sufficiently small. In contrast, the synchronous condensors and static VArsystems provide continuous control of the reactive power and can follow fast load changes.

7.1.1 Synchronous Condensors

Synchronous condensors are essentially synchronous motors operating at no load, with excitationcontrol to maintain the terminal voltage. Their advantages are as follows:

1. The availability of voltage source for commutation at the inverter even if the connectionto the ac system is temporarily interruptes. This also implies an increase in SCR as thefault level is increased. When the load supplied by the inverter is passive, the synchronouscondensor is essential for providing voltage sources for the line commutation at the inverter.

2. Better voltage regulation during a transient due to the maintenance of flux linkages in therotor windings. The effect of the armature reaction is counteracted during a transient byinduced currents in the field and armature circuits.

Synchronous condensors also have certain disadvantages. These are:

33

CHAPTER 7. REACTIVE POWER COMPENSATION 34

1. high maintenance and cost

2. possibility of instability due to the machine going out of synchronism.

The static VAr systems provide the fastest response following a disturbance. The configurationsnormally used are

1. fixed capacitor (FC) or thyristor controlled reactor (TCR)

2. thyristor switched capacitors (TSC)


Part II

University Exam Questions andAnswers

35

Chapter 8

Module I

8.1 Short Answer Questions (5 marks)

1. Compare ac and dc transmission systems. (November 2009)

2. Explain the modern trends in dc transmission. (November 2009)

3. What are the disadvantages of dc transmission? (July 2010)

4. Explain the overvoltage protection of valve. (July 2010)

5. Give the advantages, disadvantages and applications of dc transmission systems. (November2011)

6. Give a brief description about the reliability of dc transmission system. Also explain

• Energy availability

• Trasient stability

(November 2011)

7. What are the applications of dc transmission? (November 2010)

8. What do you mean by valve firing? (November 2010)

9. What are the types of dc links? (June 2012)

10. Explain valve design considerations. (June 2012)

11. Compare ac and dc transmission on the basis of economics of power transmission. (June2013)

12. List five applications of HVDC transmission. (June 2013)Answer:

(a) Long distance bulk power transmission

(b) Power transmission using underground or underwater cables (for distance > 32km)

(c) Asynchronous interconnection of AC systems operating at different frequencies or whereindependent control of systems is desired

(d) Control and stabilisation of power flows in AC ties in an integrated power system

(e) Multi-terminal HVDC system is used for interconecting three or more 3-phase ac sys-tems.

36

CHAPTER 8. MODULE I 37

8.2 Long Answer Questions

1. (a) Compare the ac and dc transmission. (10 marks)

(b) Explain the stability limits of ac power transfer. (5 marks) (July 2010)

2. (a) Explain about different valve tests. (10 marks)

(b) Discuss about valve design considerations. (5 marks) (July 2010)

3. (a) Compare ac and dc transmission. (10 marks)

(b) What are the types of dc links? (5 marks) (November 2010)

4. (a) Explain the modern trends in dc transmission. (10 marks)

(b) Write short note on choice of voltage level. (5 marks) (November 2010)

5. Compare ac and dc transmission. (15 marks) (June 2011)

6. (a) Explain about the thyristor valve design considerations. (5 marks)

(b) Discuss briefly about the different types of thyristor valve tests. (10 marks) (June 2011)

7. What are the factors which determine the relative merits of two modes (ac and dc) trans-mission system? Explain each. (15 marks) (November 2011)

8. Explain thyristor valve firing and the valve design considerations. (15 marks) (November2011)

9. (a) Explain a typical HVDC converter station. (10 marks)

(b) Differentiate between ac and dc transmission. (5 marks) (June 2012)

10. (a) Explain choice of voltage level for dc transmission. (10 marks)

(b) Write short notes on valve firing scheme. (5 marks) (June 2012)

11. (a) Draw the schematic diagram of a typical HVDC converter station and explain thefunction of each unit. (10 marks)

(b) Explain a back to back HVDC system with a eat diagram. (5 marks) (June 2013)

12. Analyse the advantages of dc transmission over ac transmission with reference to technicalperformance. (15 marks) (June 2013)


Chapter 9

Module II


1. What do you mean by a pulse number?(July 2010, November 2009)

2. Explain the characteristics of a 12 pulse converter. (July 2010)

3. For a 12 pulse converter with q = 4, s = 3, r = 1, calculate the maximum dc power andtransformer ratings (valve windings) if PIV rating of the valve is V and the rms rating is I.(November 2010)

4. Define pulse number and explain pulse converter circuit. (November 2011)

5. Explain the effect of source reactance on pulse converter circuit. (November 2011)

6. Derive the expression for average maximum dc voltage and valve utilisation factor. (June2012)

7. Write short notes on pulse number and transformer rating. (June 2012)

8. A transformer secondary line voltage to a 3 phase bridge rectifier is 345kV. Calculate the dcvoltage output with overlap angle = 15 and α = 15. (June 2013)

9. With a neat figure, write short note on converter bridge characteristics. (June 2013)


1. Explain with neat waveforms, the working of Graetz circuit. (15 marks) (July 2010)

2. Analyze in detail, the working of converters. (15 marks) (July 2010)

3. Explain the characteristics of a 12 pulse converter. (15 marks) (November 2010)

4. Explain 2 and 3 valve conduction. (15 marks) (November 2010)

5. With suitable waveforms analyze the working of Graetz circuit with and without overlap.(15 marks) (June 2011, November 2011)

6. Explain briefly about the working of bridge converter circuit in two and three valve conduc-tion modes. (15 marks) (June 2011)

7. Explain the rectifier and inverter characteristics of 6 pulse converter. (15 marks) (November2011)

38

CHAPTER 9. MODULE II 39

8. Explain Gratez circuit without overlap in detail. (15 marks) (June 2012)

9. (a) Explain 2 and 3 valve conduction (10 marks)

(b) Write short notes on the choice of converter configuration. (5 marks) (June 2012)

10. (a) With a neat circuit diagram and waveforms, explain the working of a Graetz circuit.(12 marks)

(b) Explain the term commutation group. (3 marks) (June 2013)

11. With a neat circuit diagram, analyse a Graetz circuit without overlap and derive the ex-pression for the average direct voltage, PIV, valve rating and VA rating for transformersecondary. (15 marks) (June 2013)


Chapter 10

Module III


1. Explain the basic principles of dc link control. (November 2009, November 2011)

2. Write short note on surge arrestors. (November 2009)

3. Explain about the current control and regulation of the voltage in the link. (July2010)

4. What are the different types of converter faults in general? (July 2010)AnswerAccording to the origin of the malfunction, converter faults can be divided into 3 broadgroups.

(a) Faults due to malfunction of valves and controllers

i. Arc backs (or back fires) in mercury arc valves

ii. Arc through (fire through)

iii. Misfire

iv. Quenching or current extinction

(b) Commutation failures in inverters

(c) Short circuits in a converter station

5. Explain the hierarchial control structure for a dc link. (November 2011, November 2010,June 2012)

6. Explain briefly the choice of converter configuration for any pulse number. (November 2010)

7. What are the drawbacks of IPC scheme and how can it be overcome? (November 2010)Drawbacks

(a) Aggravation of harmonic stability problem

(b) Higher magnitudes of non-characteristic harmonics (h 6= np± 1)

(c) Causes perturbations in zero crossing of voltage waveforms

(d) Non equidistant firing pulses results in non-symmetric output voltage

(e) All the above problems are aggravated at the frequencies for which the filter impedanceand the system impedance are in parallel resonance

(f) Cannot be used in weak ac systems

Theses problems due to harmonics instability can be overcome by

(a) Providing additional filters to filter out the non-characteristic harmonics.

40

CHAPTER 10. MODULE III 41

(b) Use of filters in control circuit to filter out non-characteristic harmonics in the com-mutation voltage. This can however be problematic due to variations in the supplyfrequency and will add to the control delays.

(c) Use of a firing angle control scheme which is independent of the zero crossing. Thislead to the development of EPC.

8. Write short notes on misfire and commutation failure. (June 2012)

9. Explain the heirarchial control structure for a dc link with a neat diagram. (June 2013)

10. Write short note on the various causes of over-voltages in a coverter station. (June 2013)


1. a) Explain the different methods of firing angle control. (7 marks)

b) Explain about different converter faults. (8 marks)(November 2009)

2. Discuss about the overvoltages in a converter station and the protection methods against it.(15 marks) (July 2010)

3. Explain about the starting and stopping of dc link. (15 marks) (July 2010)

4. a) A monopolar HVDC link has one bridge at each terminal. The parameters of the linkare:αmin = 5, γmin = 18, Rd = 5Ω, Rcr = 10Ω, Rci = 12Ω, Vd0r = 115kV ,Iref at the rectifier = 1kA, Iref at the inverter = 900A. Calculate Id, α, γ, Pi and Qi if

(i) Vd0i = 117.5kV

(ii) Vd0i = 120kV

(10 marks)

b) Write short notes on surge arresters. (5 marks) (November 2010)a) In order to determine Id, we need to know the operating mode which can be any ofthe three possibilities

i. Current control at the rectifier, γi = γmin

ii. Current control at the inverter, αr = αmin

iii. αr = αmin, γi = γmin

To identify the operating mode, we need to calculate

Id =Vd0r cosαmin − Vd01 cos γmin

Rcr +Rl −Rci(10.1)

If Id > Iref at the rectifier, then the operating mode is current control at the rectifier. IfId < Iref at the inverter, the operating mode is current control at the inverter. If bothconditions are not met, then the operating mode is αr = αmin, γi = γmin.

When Vd0i = 117.5kV ,Id = 937.7A

The mode of operation is αr = αmin, γi = γmin. Hence,

αr = 5

γi = 18

Vdi = Vd0i cos γmin −RciId = 100.5kV

Pi = VdiId = 92.24MW

cosφi =Vd

Vd0i= 0.855

Qi = Pi tanφi = 57.1MVAr


CHAPTER 10. MODULE III 42

When Vd0i = 120kV ,

5. Explain converter faults. (15 marks) (November 2010)

6. Explain briefly about the principles of DC link control. (15 marks) (June 2011)

7. Explain briefly about different types of converter faults. (15 marks) (June 2011)

8. Explain basic converter control characteristics. What are the two requirements which nece-assitate the modification of the control characteristics? (15 marks) (November 2011)

9. With the help of neat diagrams and waveforms, explain Individual Phase Control (IPC) andEquidistant Pulse Control (EPC). (15 marks) (November 2011)

10. Discuss the principles of DC link control and converter control characteeristics. (15 marks)(June 2012)

11. Explain firing angle control. (15 marks) (June 2012)

12. (a) Write short note on surge arrester. (5 marks)

(b) Explain the three basic types of faults that can occur in converters. (10 marks) (June2013)

13. With neat circuit diagram and waveforms, explain individual phase control and equidistantpulse control. (15 marks) (June 2013)


Chapter 11

Module IV


1. Explain the characteristics and types of DC breakers. (November 2009)

2. Write short note on smoothing reactors. (July 2010, November 2009, June 2012)

3. Explain the sources of reactive power. (July 2010, November 2009)

4. What are the technical advantages of a STATCOM over a SVC? (November 2010)Answer:

(a) Faster response (fraction of a cycle) which is independent of the network impedance.

(b) Requires lesser space as bulky passive components (such as reactors) are eliminated.

(c) Inherently modular and relocatable

(d) It can be interfaced with real power sources such as battery, fuel cell or SMES (Super-conducting Magnetic Energy Storage)

(e) A STATCOM has superior performance during low voltage condition as the reactivecurrent can be maintained constant. It is even possible to increase the reactive currentin a STATCOM under transient conditions if the devices are rated for the transientoverload.

5. Explain briefly about a dc line insulator. (November 2010)

6. What are the three types of ac filters? Explain each. (November 2011)

(a) Single tuned filter

(b) Double tuned filter

(c) High pass filter

i. Second order filter

ii. C type filter

7. Give a brief description about multi terminal dc (MTDC) system and its applications.(November 2011)

8. Explain types of ac filters. (June 2012)

9. Explain any five functions of smoothing reactors. (June 2013)Answer:

(a) To reduce the incidence of commutation failure in inverters caused by dips in the acvoltage at the converter bus.

43

CHAPTER 11. MODULE IV 44

(b) To prevent consequent commutation failures in inverters by reducing the rate of rise ofdirect current in the bridge when the direct voltage of another series conected bridgecollapses.

(c) To smoothen the ripple in the direct current in order to prevent the current frombecoming discontinuous at light loads.

(d) To decrease harmonic voltages and currents in the dc line.

(e) To limit the crest current in the rectifier due to a short circuit on the dc line.

(f) To limit the current in the valves during the converter bypass pair operation, due tothe discharge of shunt capacitances of the dc line.

10. With a neat diagram, explain the general arrangement of a HVDC circuit breaker. (June2013)


1. (a) Discuss about the protection of DC line. (5 marks)

(b) Explain the characteristics and types of DC breakers. (10 marks)

(November 2009)

2. Explain and discuss about the different functions of smoothing reactors. (15 marks) (July2010)

3. Explain in detail about static VAR system. (15 marks) (July 2010)

4. Explain dc breakers. (15 marks) (November 2010)

5. (a) What are the effects of corona? (10 marks)

(b) Explain the functions of smoothing reactors. (5 marks) (November 2010)

6. Explain about the basic concepts, types and characteristics of DC breakers. (15 marks)(June 2011)

7. Write short note on:

(a) Different types of multi-terminal DC system.

(b) Control and protection of MTDC system.

(15 marks) (June 2011)

8. Describe Thyristor Controlled Reactor (TCR) and Thyristor Switched Capacitor (TSC). (15marks) (November 2011)

9. What are the two possible types of MIDS systems? Explain and compare these two systems.(15 marks) (November 2011)

10. Describe the effects of Corona and DC line insulators. (15 marks) (June 2012)

11. Explain dc breakers in detail. (15 marks) (June 2012)

12. (a) Explain the various simulation tools that can be employed for the simulation of HVDCsystems. (10 marks)

(b) List any five requirements of a good simulation tool. (5 marks) (June 2013)Answer:b) The requirements of a good simulation tool are as follows:

i. Ease of maintenance


CHAPTER 11. MODULE IV 45

ii. Accuracy of solution

iii. Flexibility of use

iv. Reduced cost

v. Ease of setting up

vi. Real time simulation

vii. Easy monitoring and control over the simulation process

13. (a) Breafly explain the various reactive power sources at a converter bus. (5 marks)

(b) With neat circuit diagrams, explain the working of a thyristor controlled reactor andthyristor switched capacitor. (10 marks) (June 2013)


References

[1] E. W. Kimbark, Direct Current Transmission, Volume I, Wiley.

[2] K. R. Padiyar, HVDC Power Transmission Systems, Technology and System Interactions,New Age International, 2005.

[3] S. Rao, EHV-AC, HVDC Transmission & Distribution Engineering, Khanna Publishers.

[4] S. Kamakshaiah and V. Kamaraju, HVDC Transmission, Tata McGraw Hill, 2012.

46

Appendix A

Formulae

sin (α± β) = sinα cosβ ± cosα sinβ (A.1)

cos (α± β) = cosα cosβ ∓ sinα sinβ (A.2)

47

Appendix B

Derivations

B.1 Mode 2

Vd =3

π

[∫ u+α

α

−3

2ec dωt+

∫ π/3+α

u+α

ebc dωt

](B.1)

=3

π

[−3

2Em

∫ u+α

α

sin(ωt− π

2

)dωt+

√3Em

∫ π/3+α

u+α

sin(ωt+

π

3

)dωt

](B.2)

=3

πEm

[−3

2

∫ u+α

α

(sinωt cos

π

2− cosωt sin

π

2

)dωt

+√3

∫ π/3+α

u+α

(sinωt cos

π

3+ cosωt sin

π

3

)dωt

](B.3)

=3

πEm

[3

2

∫ u+α

α

cosωt dωt +

√3

2

∫ π/3+α

u+α

sinωt dωt+3

2

∫ π/3+α

u+α

cosωt dωt

](B.4)

=3

πEm

[3

2

∫ π/3+α

α

cosωt dωt +

√3

2

∫ π/3+α

u+α

sinωt dωt

](B.5)

=3

πEm

[3

2[sinωt]

π/3+αα −

√3

2[cosωt]

π/3+αu+α

](B.6)

=3

πEm

[3

2

[sin

(π3+ α

)− sinα

]−

√3

2

[cos

(π3+ α

)− cos (u+ α)

]](B.7)

=3

πEm

[3

2

[√3

2cosα+

1

2sinα− sinα

]−

√3

2

[1

2cosα−

√3

2sinα− 1

2cosα

]](B.8)

=3

πEm

[3√3

4cosα− 3

4sinα+

√3

2cos (u+ α) +

3

4sinα−

√3

4cosα

](B.9)

=3

πEm

[√3

2cosα+

√3

2cos (u+ α)

](B.10)

=3√3

2πEm [cosα+ cos (u+ α)] (B.11)

=Vd0

2[cosα+ cos (u+ α)] (B.12)

48

Index

AC filter, 32

control, 25

Filters, 31

Graetz circuit, 15Ground return, 10

Harmonics, 31high-pass filter, 32

IPC, 26

stability, 10

TUF, 12

49

Documents

EE705(B) HIGH VOLTAGE D.C. TRANSMISSION...Graetz circuits (simpli ed analysis only) - with and without overlap. Analysis of 2 & 3 valve conduction mode and 3 & 4 valve conduction mode